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cexpressionprecisionnumeric
DevBee
DevBee

Reputation: 77

C arithmetic expression inconsistency with printf and assignment to another variable

Below program outputs different values for a/100 where a is int.

void main()
{
    int a = -150;
    float f;
    f = a/100;
    printf(" %f,%f ",f,a/100);
}

Output: -1.000000,0.000000

Here second %f value comes up to 0.000000, but how is it possible, it should be -1.000000 or -1.500000.

Upvotes: 0

Views: 61

Answers (1)

Jean-François Fabre
Jean-François Fabre

Reputation: 140168

There is no inconsistency:

 f = a/100;

performs integer division stored into a float: result 1.0

then

 printf(" %f ",a/100);

prints an integer (1), but with float format: undefined behaviour.

this works without surprises:

void main()
{
    int a = -150;
    float f;
    f = a/100.0;
    printf(" %f,%f ",f,a/100.0);
}

printf is a variable arguments function (prototype: void printf(const char *,...)), which has no other choice than to "believe" the format that you pass.

If you tell printf that second argument is a float, then it will read the argument bytestream as a float, and if the data doesn't match a float structure, well, it won't work properly.

EDIT: compiler makers know that this is a common mistake, so they have added a special check for the printf-like functions: if you compile your file with gcc using -Wall or -Wformat options you'll get the following explicit message:

foo.c:8:1: warning: format '%f' expects argument of type 'double', but argument 3 has type 'int' [-Wformat=]
 printf(" %f,%f ",f,150/100);

(same goes if the number of %-like format arguments and the number of arguments do not match).

Upvotes: 4

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